Step-by-step guidelines to draw the shear force and bending moment diagrams using graphical method.

1. Shear Force and Bending
Moment Diagrams
Step-by-step Guidelines

2. 2
A cantilever beam ABCDE is subjected to the loadings as shown in Figure Q1.
Draw the shear force and bending moment diagrams of the entire beam and
indicate all important points on the diagrams.
Figure Q1

3. 75 kN
168.75 kN.m
Firstly, find the reactions at the support.
3

4. 75 kN
168.75 kN.m
V (kN)
x (m)
1.5 3.0 4.5
-ve
constant
load
-ve
constant
slope
Start
from 0
4

5. 75 kN
168.75 kN.m
V (kN)
x (m)
1.5 3.0 4.5
Total
shear
force =
-30 x 1.5
= -45 kN
- 45
5

6. 75 kN
168.75 kN.m
V (kN)
- 45
x (m)
1.5 3.0 4.5
No external
load/support
Shear force
remains constant
Note that external
moment does not
influence the
internal shear force
6

7. 75 kN
168.75 kN.m
V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
30kN
downward
30kN
7

8. 75 kN
168.75 kN.m
V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
No external
load/support
Shear force
remains constant
8

9. 75 kN
168.75 kN.m
V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
75kN
75kN upward
Back to 0
9

10. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
10
M (kN.m)
x (m)
2.4
-ve
increasing
shear force
-ve
increasing
slope

11. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
11
M (kN.m)
- 33.75
x (m)
2.4
Total shear force
= 0.5 x 1.5 x -45
= -33.75

12. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
12
M (kN.m)
- 33.75
x (m)
2.4
-ve
constant
shear
force
-ve
constant
slope

13. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
13
M (kN.m)
- 33.75
x (m)
2.4
Total shear
force
= 0.9 x -45
= -40.5
40.5kN.m
- 74.25

14. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
14
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
CW moment
Hint:
CW moment:
ADD to the BMD
CCW moment:
MINUS from the BMD
45kN.m

15. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
15
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
-ve
constant
shear
force
-ve
constant
slope

16. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
16
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
- 56.25
Total shear
force
= 0.6 x -45
= -27
27kN.m

17. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
17
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
- 56.25
-ve
constant
shear
force
-ve
constant
slope

18. V (kN)
- 45
- 75
x (m)
1.5 3.0 4.5
18
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
- 56.25
- 168.75
Total shear
force
= 1.5 x -75
= -112.5
112.5kN.m

19. V (kN)
- 45
- 75
1.5 3.0 4.5
19
M (kN.m)
- 33.75
x (m)
2.4
- 74.25
- 29.25
- 56.25
- 168.75
168.75 kN.m
75 kN
168.75kN.m
Back to 0
CW moment:
ADD to the BMD